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Xelga [282]
4 years ago
13

Pumping alone at respective constant one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fi

lls the same empty tank to 2/3 of capacity in 6 hours.
How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?
A. 3.25
B. 3.6
C. 4.2
D. 4.4
E. 5.5
Mathematics
1 answer:
4vir4ik [10]4 years ago
6 0

Answer: B. 3.6

Step-by-step explanation:

Given : Time taken by one inlet pipe to empty \dfrac{1}{2} of tank =  3 hours

⇒ Time taken by one inlet pipe to empty tank (entire) =  2 x 3 hours

= 6 hours → t_1

Time taken by  second inlet pipe to empty \dfrac{2}{3} of tank =  6 hours

⇒ Time taken by second inlet pipe to empty tank (entire) = \dfrac{3}{2}\times6= 3\times6

= 9 hours  → t_2

If both pipes are pumping simultaneously at their respective constant rates to fill the empty tank to capacity.

Then time taken by them to fill the empty tank to capacity would be  \dfrac{t_1t_2}{t_1+t_2}

=\dfrac{(6)(9)}{6+9}=\dfrac{54}{15}=\dfrac{18}{5}=3.6

Hence, it will take 3.6 hours to fill the empty tank to capacity .

Therefore , the correct answer is  B. 3.6 .

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